Re: Snap 705N/4100 lives and thrives!
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Funny, because I WAS going to say that your solution sounded a bit complex, but when you put it that way... LOL |
Re: Snap 705N/4100 lives and thrives!
Well, for a Snap 4000, it seems to me from what we have here in this thread is;
1 - Use Hard Disks using less than or equal to approx 2.25 A startup power 2 - Change the power supply 3 - Install a delay circuit to starts Hard Disks in sequence Also, since we do not want to exceed say 75% full load on the power supply during normal operation, we want to use Hard Disks that use no more than 1.125 A during normal post spin up operation for options 1 and 3 above. To summarize, since most users will not change the power supply or install a delay circuit, this means a stock Snap 4000 needs Hard Disks that use no more than 2.25 A on the 12 volt rail during spin up and no more than 1.125 A on the 12 volt rail during normal post spin up operation. So the question is, what large IDE Hard Disks meets these requirements, using REAL specs, not just default whole series specs listed by most manufacturers? |
Re: Snap 705N/4100 lives and thrives!
Most users have been able to use 250-300 gigs without any problems. These drives proably only have 2 maybe 3 platters. The next step up (4 platters, 400 gig) puts them over the edge. The new perpinduclar technology may help solve this.
You know the 2.5" notebook drives may be an option. My Hatichi 7K100 pulls 1.1 amps. But the capacity is well below that of 3.5" drives. |
Re: Snap 705N/4100 lives and thrives!
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Re: Snap 705N/4100 lives and thrives!
I had given it some thought, but now you are talking about 100-120GB a drive, so you might was well stick with 3.5's as they are typically faster than laptop drives.
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Re: Snap 705N/4100 lives and thrives!
You are correct on the speed but they are getting faster with the perpendicular technology. I think Samsung just anounced 200gig.
I wonder it we could use a current limiting resistor on the 12v to kill the inrush???? |
Re: Snap 705N/4100 lives and thrives!
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The CAP would increase the initial current draw and thus increase the original problem. BUT, you could use a current limiter (not a resistor). The drives would take a bit longer to spin up (not much I would bet), but might work as long as they are not being restricted too much (slowing their spin up too much and making the drive look like it failed). But this solution is too easy, and thus not fun... ;) |
Re: Snap 705N/4100 lives and thrives!
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A cap would cause an initial current rush, IF it was not charged ahead of time. The cap would have to be installed and then charged with a resistor in line (to keep the current rush down so the cap will not blow the power supply). Once the cap is charged, and resistor removed, then the drives could be hooked up and used as normal. The cap levels out the initial current inrush, and charges once the voltage/amperage levels return to "normal" after the drives spin up. Yeah, the current limiter WOULD be too easy :) We're looking for something that will keep us down in the basement tinkering for longer than 15 mins... |
Re: Snap 705N/4100 lives and thrives!
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I = E / R (Current equals Voltage divided by Resistance) The "R" part of the equation is increasing, thus causing the "I" part of the equation to decrease in return. Example: In our Snap Server 4000, we are looking at the 12 volt rail and it's current. While a Hard Disk is not a resistor, it is a Load, and this Load can be shown as a resistance in Ohms Law equations. In this case, we know the Voltage (E) as 12 Volts and we know we are talking about a max Current (I) of 2.25 Amps. Using Ohms Law, R = E / I, we can determine the Load Resistance as 12 / 2.25 (12 divided by 2.25). This means at spin up, the Load resistance is 5.34 Ohms. 5.34 = 12 / 2.25 Now this Load resistance is not going to change, the drive motor is what it is, period. So we want to know what will happen if we add a current limiting resistior in the circuit. Let's say just for this example, a 10 Ohm Resistor. This time we will use the equation I = E / R to find this new Current drain. The voltage is still 12 volts, but now our resistance is 15.34 Ohms. This is from adding the 10 Ohm resistor with the 5.34 Ohm Load Resistance of the drive motor at spin up (adding when in series, dividing when parallel). So we can determine this new current load as 12 / 15.34 (12 volts divided by the 15.34 total resistance). This means our new current with the current limiting resistor is .78 A, a far cry from the 2.25 we started with before. But the point is, it went down, not up. As the resistance increases, the current will decrease with the same fixed voltage source (in this case, the 4000 power supply is a fixed voltage source). Also, you said, "If you drop the voltage to a device, it will draw more current to compensate". This is also wrong, sorry. Given a Load Resistance (R), as the Voltage (E) goes down, so will the Current (I). This is shown in the same equation I = E / R of Ohms Law. Example: 12 volts and 10 Ohms. This is 12 / 10 = 1.2 which means 1.2 A current. 10 volts and 10 Ohms. This is 10 / 10 = 1 which means 1 A current. As voltage went down, so did current. See what happens when you argue electronics with an electronic engineering tech? ;) :D (all in fun, honest) NOTE: For those who may not know, and actually care, Ohms Law is a set of 12 equations used in electronics to help determine unknown values in a circuit using known values. The four values concerned are; P = Power in Watts E = Voltage in Volts I = Current in Amps R = Resistance in Ohms The primary equations are; P = I * E (Power equals Current times Voltage) E = I * R (Voltage equals Current times Resistance) The other 10 equations can be determined with basic algebra by these two base equations. The idea here is to be able to determine any one unknown value given two known values. For example, in the message above, I used the equation E = I * R to determine the equations required for the situation. I used E = I * R to derive the two equations I = E / R and R = E / I. Another example would be to determine Power (P) with the known values of Current (I) and Resistance (R). We know P = I * E and we know E = I * R. We can substitute the E in the first equation with the equal values to E of the second equation, leaving P = I * (I * R), or P = I * I * R or P = I Squared * R (sorry, I have no idea how to show a squared on this keyboard). |
Re: Snap 705N/4100 lives and thrives!
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...to be honest, I should have known better. It has been waaayyyy too long since I have had to apply ohms law to anything. If I would have thought about it for 5 seconds before posting we could have avoided this whole mess :D Hey, never any offense taken. We are all here to learn and share. Thanks for the info! |
Re: Snap 705N/4100 lives and thrives!
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You fell into a logical trap that Prof K. in EE413 would have rapped our knuckles over.... The fixed factor is the MECHANICAL power needed to spin the drive. THAT begets the electrical power needed, in watts. When you lower the available voltage, and require the same output power, the current needs go UP, not down. Reduced-volated starting is always harder on the motor than across-the-line, for that reason. [It's easier on the grid, and sometimes on the shaft load...] Where lowering the Ein works is with a dependent load, say a lamp. THERE, reducing Ein will reduce the Wout, and thus Iin as well.. |
Re: Snap 705N/4100 lives and thrives!
Exactly, and I thought about it after Phoenix took me behind the woodshed :)
Now that we have that all cleared up, what was this thread about anyway??? :D |
Re: Snap 705N/4100 lives and thrives!
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I will say this, you are correct in that the power requirment to spin up the motor according to spec has not changed, but if that power (i.e. voltage to drive the current needed to accimplish this power requirment) is not available, this does not mean it will magicly happen anywise. The motor in this case will just spin up slower than in spec, using a lower available power. Power requirements do not drive Current or Voltage. If the voltage is just not available, or if the current is just not available, then the power will not be there. It really is that simple. And for the record, there is a big difference between what happens in DC motor circuits and AC motor circuits (hint hint). But again, this IS NOT worth arguing over. If you think it will work, go ahead and try it, and please be sure to post your results. (even when you find out I was right, LOL, heh heh) :D |
Re: Snap 705N/4100 lives and thrives!
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Then it made a short detour into the land of, "how do we get around current requirements of the largest drives"? Which leads to discussions on electronic theory. Hey! You asked... Oh and by the way, yes, I am a smart ass. :D LOL, gotta have fun in life, right? (my apologies to the admin for the use of that one bad word, I could not figure a way around it) |
Re: Snap 705N/4100 lives and thrives!
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Re: Snap 705N/4100 lives and thrives!
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He'd filter it if he did |
Re: Snap 705N/4100 lives and thrives!
So what we have ended up with in this thread is....upgrade the power supply or get yourself some low power drives. We have overlooked one valuable avenue to solve the problem:
Why not wire up one of those little 12V bicycle generators, attatched to a stationary bike, to the power supply? Just peddle your butt off until the drives spin up! Problem solved, and it also helps with the sometimes sedentary lifestyle of the techie/geek. |
Re: Snap 705N/4100 lives and thrives!
This thread is gone down hill, but it has been fun.
There nothing says you can't use a external power power supply. As long as the grounds are tied together for reference it should work. Put all of the drives off the external. Buy a 8 bay drive unit w/ps then you expand the unit with a 2 drive ribbon connector. Expanding the raid 5 array up to 7-8 drives ( 1 spare). 7 x 300 = 2.1 T - overhead may be 2 tera bytes. How sweet would this be ......... |
Re: Snap 705N/4100 lives and thrives!
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Yes, this is the right approach. Attempts to limit the starting surge will not solve the bigger issue that you really do NOT want to screw around with marginal power supplies, period. Murphy will bite you on the *ss for doing so. |
Re: Snap 705N/4100 lives and thrives!
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So back on topic (at least part of it)... Are we agreed that in a 4100, the power supply is capable of 4 x 160 drives, which already exceeds the LBA48 limit on them? So no problems there (yet)? -AND- Are we agreed, that 1) without adding in a new delay or current limiting circuit, or 2) changing the power supply in some manner, that a 4000 should handle any version of drives x 4 so long as they require no more than; 1) 2.25 Amps Spin-Up power (9 Amps Total) AND 2) 1.125 Amps normal operating power (4.5 Amps Total) on the 12 volt rail? If, and I repeat IF, we are agreed, then what are the largest drives from the different manufacturers that will meet these requirements? (I am assuming we should always be good on the 5 volt rail) The catch to this question above is finding actual requirements of a given drive versus what the manufacturers post on their web pages. Even Storage Review has very limited info in this area (the number of drives listed). I think in this case, we are gonna be back to either doing a crap load of measurments on drives we don't have, or finding a source of people who have changed drives and sort out what has and has not worked (not just drive size, but manufacturer and model number drive). Maybe, just maybe, we can do a little SWAG (Scientific Wild A#$ed Guess) and just guess based on the number of platters used in a drive versus the number of platters used in upgrades we know worked. This last one would probably preclude Seagates because they seem to draw a lot more power on the 12 volt rail than other manufacturers. Any ideas? |
Re: Snap 705N/4100 lives and thrives!
WRT Larger drive I have had 4 x 250Mb Quantum 2500BB drives running in a 4100 without problem. The big issue is start up current. I have noticed that my 4100 DOES NOT spin-up all the drives at the same time, but each one comes on after a short delay (app 10 secs) after the previous one. Maybe there is a BIOS tweek that could lengthen the delay as the bigger drive do take longer to come up to speed. The other thing to check is not the total power budget, but the power requirements on each supply voltage as it is likely to be the 12V supply that is pulling too much current. If so an additional 12V onlt supply may give the extra necessary to keep the supply up during start-up. Some drive can pull 2-4 Amps on the 12v supply for the first few seconds, that's a lot of (peak) power worst case over 200W plus the 5v supply of perhaps another 50W .......
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Re: Snap 705N/4100 lives and thrives!
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Re: Snap 705N/4100 lives and thrives!
Looking at the debug cmds for a 2200, it does not give any options for delay or stager start.
Need to have someone (jontz) look at all of the debug cmds for the 4100 or someone with a 4000 to see if its a opton. I know different version of the OS have different option, I always assume it was the same for all platforms. |
Re: Snap 705N/4100 lives and thrives!
I looked through all of the debug commands in my 4100, there is no option to either enable or disable the delay startup routine. I'll keep digging though...
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